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Difficulty: MediumSolubility Curves and Temperature Effects

The solubility of lead(II) trioxonitrate(V), Pb(NO3)2\text{Pb(NO}_3\text{)}_2, in water is 3.5 mol dm−33.5\text{ mol dm}^{-3} at 80∘C80^\circ\text{C} and 1.5 mol dm−31.5\text{ mol dm}^{-3} at 30∘C30^\circ\text{C}. What mass of Pb(NO3)2\text{Pb(NO}_3\text{)}_2 will crystallize out when 250 cm3250\text{ cm}^3 of its saturated solution is cooled from 80∘C80^\circ\text{C} to 30∘C30^\circ\text{C}? [Molar mass of Pb(NO3)2=331 g mol−1\text{Pb(NO}_3\text{)}_2 = 331\text{ g mol}^{-1}]

  1. 165.5 g165.5\text{ g}Answer
  2. B
    0.50 g0.50\text{ g}
  3. C
    662.0 g662.0\text{ g}
  4. D
    289.6 g289.6\text{ g}

Answer

The mass of Pb(NO3)2\text{Pb(NO}_3\text{)}_2 that will crystallize out is 165.5 g165.5\text{ g}.
Subtracting the molar solubility at 30∘C30^\circ\text{C} from that at 80∘C80^\circ\text{C} gives a precipitation rate of 2.0 mol dm−32.0\text{ mol dm}^{-3}. Multiplying by the molar mass (331 g mol−1331\text{ g mol}^{-1}) gives 662.0 g662.0\text{ g} per dm3\text{dm}^3. Scaling this value for 250 cm3250\text{ cm}^3 (0.25 dm30.25\text{ dm}^3) yields 165.5 g165.5\text{ g}.

Step-by-Step Solution

1
Calculate the difference in molar solubility between 80∘C80^\circ\text{C} and 30∘C30^\circ\text{C}.
Molar solubility difference=3.5 mol dm−3−1.5 mol dm−3=2.0 mol dm−3\text{Molar solubility difference} = 3.5\text{ mol dm}^{-3} - 1.5\text{ mol dm}^{-3} = 2.0\text{ mol dm}^{-3}
This determines the amount of solute in moles that precipitates out per dm3\text{dm}^3 of solvent.
2
Calculate the mass of solute that crystallizes out from 1 dm31\text{ dm}^3 (1000 cm31000\text{ cm}^3).
Mass crystallized in 1000 cm3=2.0 mol dm−3×331 g mol−1=662.0 g dm−3\text{Mass crystallized in } 1000\text{ cm}^3 = 2.0\text{ mol dm}^{-3} \times 331\text{ g mol}^{-1} = 662.0\text{ g dm}^{-3}
Converting the solubility difference from moles to grams using the given molar mass.
3
Scale the mass to the given solution volume of 250 cm3250\text{ cm}^3.
Mass crystallized=662.0 g×250 cm31000 cm3=165.5 g\text{Mass crystallized} = 662.0\text{ g} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3} = 165.5\text{ g}
The volume provided (250 cm3250\text{ cm}^3) is one-quarter of 1000 cm31000\text{ cm}^3 (1 dm31\text{ dm}^3).

Key Concept

Mass of solute precipitated upon cooling a saturated solution
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Topics

Solubility Curves and Temperature Effects

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